problem: Given $X=\left\{ \begin{pmatrix} m & n \\ n & m \end{pmatrix} \mid m,n \in \Bbb Z \right\}$, find all the maximal ideal of $X$.

This's my ideas: let $f: X \to Z$x$Z$ , $\begin{pmatrix} m & n \\ n & m \end{pmatrix} \to (m,n) $. f is bijective function, so $X=Z$x$Z$

If $K$ is ideal of $X$, $K=I$x$J$, $I,J$ are ideal of $Z$.

$Z$ is domain ring, so $K=mZ$x$nZ$

$K$ is max ideal iff just one $I$ or $J$ is max ideal (Since product of two integral domains is not an integral domain).

so $K=A=\left\{ \begin{pmatrix} pk & b \\ b & pk \end{pmatrix} \mid k,b \in \Bbb Z \right\}$ or $K=B\left\{ \begin{pmatrix} a & pk \\ pk & a \end{pmatrix} \mid a,k \in \Bbb Z \right\}$ ($p$ is prime number).

but now $A,B$ are not ideal of $X$.

Where is my mistake? Help me sholve this.

Thank you very much.

  • $\begingroup$ what is the operation defined on $\times$ in $K = I \times J$ ? $\endgroup$ – onurcanbektas Jun 25 '17 at 16:30
  • $\begingroup$ $I=\left\{ m \in \Bbb Z: \begin{pmatrix} m & n \\ n & m \end{pmatrix}\in K \right\}$ $\endgroup$ – Chloe.Sannon Jun 25 '17 at 16:52
  • $\begingroup$ $J=\left\{ n \in \Bbb Z: \begin{pmatrix} m & n \\ n & m \end{pmatrix}\in K \right\}$ $\endgroup$ – Chloe.Sannon Jun 25 '17 at 16:55
  • $\begingroup$ No, I mean what is the definition of the operation between the ideals ? I'm asking this because in the equation $K = I \times J$, the LHF is matrix, and the RHS is 2-tuples(probably), or something, but definitely not matrix. $\endgroup$ – onurcanbektas Jun 25 '17 at 17:12
  • $\begingroup$ @AnhThư You are mistaken when assume that $X=\mathbb Z\times\mathbb Z$ (actually this is an isomorphism) since you consider $\mathbb Z\times\mathbb Z$ as a ring with multiplication $(a_1,a_2)(b_1,b_2)=(a_1b_1,a_2b_2)$ (only in this case can decide that an ideal is of the form $I\times J$). $\endgroup$ – user26857 Jun 28 '17 at 22:54


An ideal $I$ of a unitary commutative ring R is maximal iff $R / I$ is a field.

  • $\begingroup$ Thanks, i've known it.but I don't know how to find they $\endgroup$ – Chloe.Sannon Jun 25 '17 at 16:50

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